The weight w of an object varies inversely as the square of the distance d from the center of the earth. At sea level (3978 miles from the center of the earth), an astronaut weighs 110 pounds. Find her weight when she is 139 miles above the surface of the earth and the spacecraft is not in motion.
I know that this is supposed to be set up as a proportion but i'm not exactly sure how to do so because I have set it up every way possible and I still don't get the right answer.
See previous post: Tue, 5-12-09, 9:00 PM.
To solve this problem, we can use the inverse square proportion:
w = k/d^2
where w is the weight, d is the distance from the center of the Earth, and k is a constant of variation.
The problem gives us the weight (w) at a specific distance (d) from the center of the Earth, which allows us to find the constant of variation (k).
At sea level, the weight (w) is 110 pounds, and the distance (d) is 3978 miles:
110 = k/3978^2
Now we can solve for k:
k = 110 * 3978^2
Once we have the value of k, we can use it to find the weight (w) at a different distance (d).
Given that the distance (d) is 139 miles above the surface of the Earth:
w = k/d^2
w = 110 * (139)^2
Calculating this will give us the weight (w) of the astronaut when she is 139 miles above the surface of the Earth.